Problem: $f'(x)=-6f(x)$, and $f(2)=1$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $f(x)=e^{-6x}$ (Choice B) B $f(x)=-12e^{-6x}$ (Choice C) C $f(x)=2e^{-x}$ (Choice D) D $f(x)=e^{12-6x}$
The general solution of equations of the form $f'(x)=kf(x)$ is $f(x)=C\cdot e^{kx}$ for some constant $C$. This can be found using separation of variables. In our case, $k=-6$, so $f(x)=C\cdot e^{-6x}$. Let's use the fact that $f(2)=1$ to find $C$ : $\begin{aligned} f(x)&=C\cdot e^{-6x} \\\\ f(2)&=C\cdot e^{-6\cdot 2} \gray{\text{Plug }x=2} \\\\ 1&=C\cdot e^{-6\cdot 2} \gray{f(2)=1} \\\\ e^{12}&=C \end{aligned}$ In conclusion, $f(x)=e^{12-6x}$.